Explore Fraction Concepts with Examples and Questions

Un-Simplifying a fraction, i.e. computing partial fraction decomposition
$\frac{3x^2+17x}{x^3+3x^2+-<!--\; -\; -->6x-<!--\; -\; -->8}$
I need to find the value of C in the form of $\frac{A}{x+1}+\frac{B}{x-<!--\; -\; -->2}+\frac{C}{x+4}$ which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-<!--\; -\; -->8)+B(x^2+5x+4)+C(x^2-<!--\; -\; -->x-<!--\; -\; -->2)=3x^2+17x$
No clue on my next step or even if this is the right step.

What is the name for the mathematical property involving addition or subtraction of fractions over a common denominator?
For any number $x$ where $x\in <!--\; \in \; -->\mathbb{R}$ and where $x\ne <!--\; \ne \; -->0$, what is the mathematical property which states that:
$\frac{1-<!--\; -\; -->x^2}{x}=\frac{1}{x}-<!--\; -\; -->\frac{x^2}{x}$

Prove this inequality:$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge <!--\; \ge \; -->\sqrt{a}+\sqrt{b}+\sqrt{c}+3$
I tried various methods. But, couldn't solve it. It'd be great if anyone can help.

Combined Probability
$P(A)\frac{2}{3},P(A|B)=\frac{1}{3}$
and
$P(A\cup B)=\frac{4}{5}.$
Find P(B).
I honestly have no idea how to even approach this problem, as I cannot find any helpful online notes on Combined probability. For instance, I don't even know how to find P(A ⋂ B) if given P(A ∪ B), P(A) and P(B).
So how would I solve a simple problem like finding P(A ⋂ B), and then how would I use this knowledge to solve the question mentioned above?

Removing a fraction in the denominator
What steps do I need to take to simplify $\frac{1}{\frac{1}{\sqrt{2}}}$ to $\frac{\sqrt{2}}{1}$? Can you please explain the steps i need to take in this problem and explain generally how to remove fractions from the denominator.

The expression $(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ where $q\ne <!--\; \ne \; -->1$, equals
(A) $\frac{1-<!--\; -\; -->q^{128}}{1-<!--\; -\; -->q}$
(B) $\frac{1-<!--\; -\; -->q^{64}}{1-<!--\; -\; -->q}$
(C) $\frac{1-<!--\; -\; -->q^{2^{1+2+\cdots <!--\; \cdots \; -->+6}}}{1-<!--\; -\; -->q}$
(D) none of the foregoing expressions
What I have done
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ is a polynomial of degree 127. Now the highest degree of the polynomial in option (A), (B) and (C) is 127, 63 and 41 respectively. And therefore (A) is the correct answer.
I get to the correct answer but I don't think that my way of doing is correct. I mean what if, if option (B) was $\frac{1+q^{128}}{1-<!--\; -\; -->q}$
Please show how should I approach to the problem to get to the correct answer without any confusion.

Prove the inequality
$\frac{1}{1+a+b}+\frac{1}{1+c+b}+\frac{1}{1+a+c}+a+b+c\le <!--\; \le \; -->3+\frac{1}{3}(ab+bc+ac)$
I used Buffalo way. Can anyone offer an easier way

How do I simplify this expression about factorization?
I am trying to simplify this
$\frac{9x^2-<!--\; -\; -->x^4}{x^2-<!--\; -\; -->6x+9}$
The solution is
$\frac{-<!--\; -\; -->x^2(x+3)}{x-<!--\; -\; -->3}=\frac{-<!--\; -\; -->x^3-<!--\; -\; -->3x^2}{x-<!--\; -\; -->3}$
I have done
$\frac{x^2(9-<!--\; -\; -->x^2)}{(x-<!--\; -\; -->3)(x-<!--\; -\; -->3)}=\frac{x^2(3-<!--\; -\; -->x)(3+x)}{(x-<!--\; -\; -->3)(x-<!--\; -\; -->3)}$
but I do not find a way to simplify
How can I simplify to get the answer?
I have to use
$(a+b)(a-<!--\; -\; -->b)=a^2-<!--\; -\; -->b^2$
$(a+b{)}^2=a^2+2ab+b^2$
$(a-<!--\; -\; -->b{)}^2=a^2-<!--\; -\; -->2ab+b^2$

What is the best way of resolving an expression with square roots in denominator?
I was resolving the following question from my textbook:
Write each of the following expressions as a single fraction, simplifying your answer where possible:
$4-<!--\; -\; -->\frac{1}{\sqrt{12}}+\frac{10}{\sqrt{3}}$
I have resolved it this way:
$\frac{4\sqrt{12}\sqrt{3}}{\sqrt{12}\sqrt{3}}-<!--\; -\; -->\frac{\sqrt{3}}{\sqrt{12}\sqrt{3}}+\frac{10\sqrt{12}}{\sqrt{12}\sqrt{3}}=\frac{4\sqrt{36}-<!--\; -\; -->\sqrt{3}+10\sqrt{4}\sqrt{3}}{\sqrt{36}}=\frac{24-<!--\; -\; -->\sqrt{3}+20\sqrt{3}}{6}$
while in the textbook it is resolved like this:
$4-<!--\; -\; -->\frac{1}{\sqrt{4\times <!--\; \times \; -->3}}+\frac{10}{\sqrt{3}}=4-<!--\; -\; -->\frac{1}{2\sqrt{3}}+\frac{10}{\sqrt{3}}=\frac{8\sqrt{3}-<!--\; -\; -->1+20}{2\sqrt{3}}=\frac{8\sqrt{3}+19}{2\sqrt{3}}$
I am wondering did I do it correctly and if so, which way is better?

Prove if $f(x)=\ln <!--\; \; -->(1-<!--\; -\; -->\frac{1}{x^2})$ then $f(2)+f(3)+f(4)=\ln <!--\; \; -->(5/8)$
I have that:
$f(x)=\ln <!--\; \; -->(1-<!--\; -\; -->\frac{1}{x^2})$
I need to prove that $f(2)+f(3)+f(4)=\ln <!--\; \; -->\left(\frac{5}{8}\right)$
Indeed, I proved that $f(2)+f(3)+f(4)=\ln <!--\; \; -->(3/4)+\ln <!--\; \; -->(8/9)+\ln <!--\; \; -->(15/16)$ But couldn't derive $\ln <!--\; \; -->(5/8)$ using it.
Any hint is welcome!\

choosing N object from set of M>N maximizing overall ratio value/weight
I have a set of M objects each with a certain value and weight. From this set I want to take out N objects ($N<M$ of course) and maximize the ratio:
total value of the N objects / total weight of the N objects
It's a kind of an optimal selection problem.
My intuition tells me to choose the N objects with the highest value/weight ratio, that is sort the M objects according to this ratio in descending order and take the first N objects in the sorting sequence.
How can I demonstrate doing this will always yield the maximum total ratio?
My attempt: if optimal ratio is $\frac{a}{b}$ for a certain selection of $K$ objects and I have to add a $K+1$ object, between one with a ratio $\frac{x}{y}$ and another with ratio $\frac{w}{z}$, given that
$\frac{x}{y}>\frac{w}{z}$
is the former always an optimal choice?
I thought the problem could algebrically be reduced to this: given $a,b$ positive constants, and $x,y,w,z$ all positive so that:
$\frac{x}{y}>\frac{w}{z}$
can or can never be that:
$\frac{a+x}{b+y}<\frac{a+w}{b+z}$
I am not able to answer this point, please help!

Clarification about percentage calculus
Why if we want to know what percentage of 16 is 4 we do
4/16
and not
16/4
?
4/16 gives you the answer, because it's equal to 0.25, that is equal to 25%, while 16/4 doesn't (I guess); but I don't understand the logic. It seems more natural to me to do 16/4, because if I want to know the percentage I would split the 16 by 4 to know how many parts does the 4 create (in my logic the number of parts created define the percentage). So if I try to do 16/4 I get 4, but then I don't know how to go on to get 25%. Maybe there's a way to obtain the result in the way I'm proposing but I don't know it.
I hope my question doesn't sound stupid, but this is one of the many basic things that I'm lacking in math to fully understand its logic.

Evaluating limit using logarithms.
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\ln <!--\; \; -->x{)}^{\frac{1}{x}}$
What i have tried:
$\ln <!--\; \; -->[\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\ln <!--\; \; -->x{)}^{\frac{1}{x}}]$
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\ln <!--\; \; -->(\ln <!--\; \; -->x{)}^{\frac{1}{x}}$
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(\ln <!--\; \; -->x)}{x}$
So as $x$ approaches infinity, the limit goes to 0. But the answer in the book is 1.

simple integration artimethic error
I am trying to integrate a polynomial but I couldn't get the correct answer somehow. I feel like I'm making a mistake when evaluating the integral.
$\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_{-<!--\; -\; -->1}^{1}1-<!--\; -\; -->2x^2+x^{4}dx=[x-<!--\; -\; -->\frac{2x^3}{3}+\frac{x^5}{5}{]}_{-<!--\; -\; -->1}^1$
$=1-<!--\; -\; -->\frac{2}{3}+\frac{1}{5}+1-<!--\; -\; -->\frac{2}{3}+\frac{1}{5}$
$=2+\frac{2(-<!--\; -\; -->2)}{3}+\frac{2(1)}{5}$
$=\frac{30-<!--\; -\; -->26}{15}=\frac{4}{15}\mathrm{\pi <!--\; \pi \; -->}$
But the integral calculator is giving me: $\frac{16}{15}\mathrm{\pi <!--\; \pi \; -->}$

Problem in solving an Integral.
I'm solving a solid of revolution problem and I'm stuck at this point. The u-subtitution doesn´t work, I don´t know what method use.
$\mathrm{\pi <!--\; \pi \; -->}\int <!--\; \int \; -->(\frac{4x-<!--\; -\; -->1}{8x^4}{)}^2$

Proof of $\frac{q_n}{q_{n-<!--\; -\; -->1}}=[a_n,a_{n-<!--\; -\; -->1},a_{n-<!--\; -\; -->2},...,a_2,a_1]$?
Proof of continued fractions axiom.
Let $c=[a_0,a_1,a_2,\dots <!--\; \dots \; -->,a_n,\dots <!--\; \dots \; -->]=a_0+\frac{1}{a_1+\frac{1}{a_2+\ddots <!--\; \ddots \; -->}}$ be a continued fraction which could be finite or infinite.
By $\frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e.
$\frac{p_n}{q_n}=[a_0,a_1,a_2,\dots <!--\; \dots \; -->,a_n].$
How to prove the formula formula:
$\frac{q_n}{q_{n-<!--\; -\; -->1}}=[a_n,a_{n-<!--\; -\; -->1},a_{n-<!--\; -\; -->2},...,a_2,a_1]?$
This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula
$\frac{p_n}{p_{n-<!--\; -\; -->1}}=[a_n,a_{n-<!--\; -\; -->1},\dots <!--\; \dots \; -->,a_1,a_0].$

Simplification imaginary fractions
In an exercise, a partial fraction expansion has to be done. I have no problem with that, but one of the last steps includes a simplification as follows:
$$\begin{gathered} (-<!--\; -\; -->\frac{1}{2}-<!--\; -\; -->\frac{1}{6}i)\left(\frac{1}{s+1+3i}\right)+(-<!--\; -\; -->\frac{1}{2}+\frac{1}{6}i)\left(\frac{1}{s+1-<!--\; -\; -->3i}\right) \\ =-<!--\; -\; -->\frac{s+1}{(s+1{)}^2+3^2}-<!--\; -\; -->\frac{1}{3}\left(\frac{3}{(s+1{)}^2+3^2}\right). \end{gathered}$$
My problem is that I do not understand the steps in between these two results. Could anyone explain how this is done? Any help is appreciated!

Can ratios really be manipulated as fractions?
In high-school Maths, we were taught that it was possible to manipulate ratios as fractions. For example,
$$\begin{gathered} 1:7=3:x \\ \frac{1}{7}=\frac{3}{x} \\ \frac{x}{7}=3 \\ x=3\times <!--\; \times \; -->7 \\ x=21 \end{gathered}$$
My question relates to the second line of working, where we represent $1:7$ as $\frac{1}{7}$. As I understand it, $1:7$ s better represented as one part of 8 total parts, i.e. $\frac{1}{8}$ My question, then, is this:
Is there a simple way of showing (or better still, a rigorous proof!) that a ratio-equation of the form $a:n=b:m$ can in all cases be rewritten as $\frac{a}{n}=\frac{b}{m}$?

How Do I Simplify $1-<!--\; -\; -->\frac{1}{2^{m-<!--\; -\; -->1}}+\frac{1}{2^m}$
Someone help. Somehow my textbook says that it simplifies to $1-<!--\; -\; -->\frac{2}{2^m}+\frac{1}{2^m}$ I don't see this at all.

How to position negative sign of fraction
For example we have:
$\frac{-<!--\; -\; -->1}{2}$
Does this mean that only the numerator of the fraction is negative?
Can we put it like this?
$-<!--\; -\; -->\frac{1}{2}$
Does this means that the whole fraction is negative?
And then can we put it like this?
$\frac{1}{-<!--\; -\; -->2}$
Does this mean that only the denominator of the fraction is negative?
Do the above suggestions [look] right? Or how should I understand when in some examples the minus sign is moving from numerator to the whole fraction?